Question 1

Two moving particles will COLLIDE if the position vector of one relative to the other is:

Accepted Answer

Always parallel (anti-parallel) to their relative velocity. Collision condition: r⃗_rel stays along v⃗_rel (the relative velocity points from one particle straight at the other). Speed-trick: in the frame of one particle, the other must head directly at it.

Question 2

In the reference frame of particle B, particle A (both moving uniformly) appears to move:

Accepted Answer

In a straight line with constant velocity v⃗_A − v⃗_B. With both velocities constant, the relative velocity is constant, so the relative trajectory is a straight line.

Question 3

The time of closest approach for two uniformly moving particles with initial separation r⃗₀ and relative velocity v⃗_rel is:

Accepted Answer

t = -\dfrac{\vec{r}_0 \cdot \vec{v}_{rel}}{|\vec{v}_{rel}|^2}. Minimising |r⃗₀ + v⃗_rel t|² gives t = −(r⃗₀·v⃗_rel)/|v⃗_rel|². (t = r₀/v_rel only when they head straight at each other.)

Question 4

The minimum (closest-approach) distance between two uniformly moving particles equals:

Accepted Answer

The component of the initial separation perpendicular to the relative velocity. In the relative frame the moving particle travels a straight line; the closest distance is the perpendicular from the other particle to that line: d_min = r₀ sinα.

Question 5

An aeroplane's velocity relative to the ground equals:

Accepted Answer

Its velocity relative to the air PLUS the wind velocity. v⃗_plane,ground = v⃗_plane,air + v⃗_air,ground — the wind adds vectorially to the plane's airspeed.

Question 6

For a plane to fly due north when a wind blows from the west, the pilot must head:

Accepted Answer

Slightly west of north (into the wind). The heading must have a westward component to cancel the eastward push of the wind (from the west means blowing towards the east).

Question 7

With a tailwind (wind along the direction of flight), the ground speed of a plane is:

Accepted Answer

Airspeed + wind speed. A tailwind adds directly: v_ground = v_air + v_wind; a headwind subtracts.

Question 8

A stone dropped from a uniformly moving train, as seen by a passenger IN the train, falls:

Accepted Answer

Straight down (vertical straight line). Relative to the train the stone has zero horizontal velocity, so it falls vertically. To a ground observer it is a parabola. Relative trajectory depends on the frame!

Question 9

Two projectiles are in flight simultaneously (same g). The trajectory of one as seen from the other is a:

Accepted Answer

Straight line. Relative acceleration = g − g = 0, so the relative velocity is constant and the relative path is a straight line.

Question 10

For two ships to avoid collision, an observer on one ship must see the other ship's bearing (direction):

Accepted Answer

Changing with time. Constant bearing + closing range = collision course (r⃗_rel parallel to v⃗_rel). A changing bearing means they will pass clear — the mariner's rule.

Question 11

A plane's airspeed is 200 km/h due north with a 50 km/h tailwind from the south. Its ground speed is:

Accepted Answer

250 km/h. Wind from the south blows northward (tailwind): v_ground = 200 + 50 = 250 km/h.

Question 12

A plane flies at airspeed 200 km/h due north while a 50 km/h wind blows towards the east. Its ground speed is:

Accepted Answer

≈ 206 km/h. Perpendicular vectors: v = √(200² + 50²) = √42500 ≈ 206 km/h, tilted east of north.

Question 13

A plane with airspeed 100 m/s must fly due north against a 60 m/s wind blowing towards the east. The pilot must head west of north by an angle:

Accepted Answer

sin⁻¹(0.6) ≈ 37°. The westward component must cancel the wind: 100 sinθ = 60 ⇒ sinθ = 0.6 ⇒ θ ≈ 37° west of north.

Question 14

For the previous plane (100 m/s airspeed, 60 m/s crosswind cancelled), the resultant ground speed due north is:

Accepted Answer

80 m/s. v_north = √(100² − 60²) = √6400 = 80 m/s (6-8-10 triangle).

Question 15

Ship A is 10 km due east of ship B. A sails west at 6 km/h and B sails east at 4 km/h (towards each other). They will meet after:

Accepted Answer

1 h. Closing speed = 6 + 4 = 10 km/h; t = 10/10 = 1 h. (Head-on: r⃗₀ ∥ v⃗_rel, so they collide.)

Question 16

Two particles are 100 m apart. One moves at 10 m/s directly towards the other, which moves at 6 m/s directly away. The first catches the second after:

Accepted Answer

25 s. v_rel = 10 − 6 = 4 m/s (closing); t = 100/4 = 25 s.

Question 17

Ship A is 10 km north of ship B. A sails east at 3 m/s; B sails north at 4 m/s. The magnitude of B's velocity relative to A is:

Accepted Answer

5 m/s. v⃗_BA = v⃗_B − v⃗_A = (0,4) − (3,0) = (−3, 4); |v_BA| = 5 m/s.

Question 18

Two straight roads cross at right angles. Car A approaches the crossing from the west at 30 m/s, currently 400 m away; car B approaches from the south at 40 m/s, currently 300 m away. Their separation is decreasing. The relative speed of A with respect to B is:

Accepted Answer

50 m/s. v⃗_rel = (30, 0) − (0, 40) = (30, −40); |v_rel| = √(900 + 1600) = 50 m/s.

Question 19

For the same two cars (A: 400 m west of crossing at 30 m/s; B: 300 m south at 40 m/s), both reach the crossing at t = 400/30 ≈ 13.3 s and t = 300/40 = 7.5 s respectively. Do they collide at the crossing?

Accepted Answer

No — they reach the crossing at different times. A arrives at 13.3 s, B at 7.5 s — different times, so no collision at the crossing. Pitfall: intersecting PATHS do not mean collision; the times must match too.

Question 20

Particle B is initially 100 m due east of A. A moves east at 8 m/s; B moves east at 3 m/s. Their closest approach (and when) is:

Accepted Answer

0 m, at t = 20 s (A catches B). v_rel = 8 − 3 = 5 m/s along the line joining them (r⃗₀ ∥ v⃗_rel) ⇒ they meet: t = 100/5 = 20 s, d_min = 0.

Closest Approach & Wind-Airplane Problems — NEET Physics MCQs with Solutions

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